1. A solution is made by dissolving 1.13g of aluminum chloride in enough water t
ID: 906684 • Letter: 1
Question
1. A solution is made by dissolving 1.13g of aluminum chloride in enough water to make 500 mL of solution . What is the molarity of this solution?
2. Consider the following equation: 4PH3(g) ----- P4(s) + 6 H2(g) A sample of PH3(g) has an initial mass of 15.67 g. Assume all of the reactant is converted to products, and none is lost, what is the mass (in grams) of P4(s) is produced by the reaction.
7. A salt with the formula MA2 is added to water and partially dissolves. Once the equilibrium is established, [M+2] = 2.2 x 10-4M. Calculate the value of Ksp for the dissolution of the salt.
8. Calculate the solubility of BaSO4 in a 0.71M solution of Na2SO4. Ksp BaSO4 = 1.1 x 10-10.
Explanation / Answer
1.Molarity = Moles of solute dissolved in 1 liter of solution
Mass of AlCl3= 1.13g
Molecular weight of AlCl3= 27+3*35.5 =133.5 mole= mass/ molecularweight= 1.13/133.5 =0.008464 moles in 500ml solution in 1000ml (1 liter ) =0.008464*2= 0.016929 moles per liter, molarity = 0.016929M
2. The reaction is 4PH3 (g) ----> P4(s)+ 6H2 (g)
Molecular weightr of PH3= 34 gms
from the reaction 4*34=136 gms of PH3 gives 4*31=124 gms of P4
7. For a salt MA2 dissociation is given by
MA2---> M+2 + 2A-
Given [M+2] =2.2*10-4 M
[A] =2*2.2*10-4 M =4.4*10-4M
Ksp = [M+2] [A-]2 = 2.2*10-4 * (4.4X10-4)2 =4.25*10-11
8. The dissociation of BaSO4 can be represented as
BaSO4---> Ba+2 + SO4-2, Ksp =1.1* 10-10
KSp = [Ba+2] [SO4-2]
let there be change in x leading to equilibrium
at Equilibrium [Ba+2] =x and [SO4-2] =0.71+x ( since there is 0.71M Na2SO4 which suppliments SO4-2
x*(0.71+x)= 1.1*10-10
0.71x+x2= 1.1*10-10
x2+0.71x- 1.1*10-10= 0 This is a quadratic equation of the form ax2+bx+c= 0
The solution is x =1.5493*10-10
136 gms of PH3 gives 124 gms of P4
15.67 gms give 15.67*124/136 gms of P4=14.28 gms of P4.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.